Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could just split the collection in half. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets the same total value.
Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles.
Input
Each line in the input describes one collection of marbles to be divided. The lines consist of six non-negative integers n1, n2, …, n6, where ni is the number of marbles of value i. So, the example from above would be described by the input-line “1 0 1 2 0 0’’. The maximum total number of marbles will be 20000.
The last line of the input file will be ``0 0 0 0 0 0’’; do not process this line.
Output
For each colletcion, output Collection #k:’’, where k is the number of the test case, and then either ``Can be divided.’’ or Can’t be divided.’’.
Output a blank line after each test case.
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1059
这道题的意思是输入6个数字，分别代表价值为1，2，3，4，5，6的大理石的数量，问能不能将所有大理石分为价值相同的两份。并且不能将大理石割开。
思路：典型的多重背包问题，可以理解为在一个背包在不超过全部大理石的价值总量的1/2的情况下，能拿走多少价值的大理石，如果刚好能拿走一半，就输出Can be divided.否则就输出Can’t be divided.所以，这题既大理石的价值当作价值，也当作重量。
#include <iostream>
#include<algorithm>
using namespace std;
int mo[20010],dp[120005],b[7],val[20010];
int main()
{
int ca = 1;
while (cin >> b[1] >> b[2] >> b[3] >> b[4] >> b[5] >> b[6])
{
if (b[6] ==0&& b[1] == 0&&b[2] == 0&&b[3] ==0 &&b[4] == 0&&b[5] == 0)
break;
memset(mo, 0, sizeof(mo));
memset(dp, 0, sizeof(dp));
int sum = 0 ,k=0;
for (int i = 1; i <= 6; i++)
{
sum = sum + b[i] * i;
for (int j = 1; j <= b[i]; j <<=1)
{
val[k++] = j * i;
b[i] = b[i] - j;
}
if (b[i] > 0)
val[k++] = b[i] * i;
}
if (sum % 2 == 1)//如果总价是的奇数，肯定不能平均分。
{
cout << "Collection #" << ca++ << ":" << endl;
cout << "Can't be divided." << endl << endl;;
}
else
{
dp[0] = 0;
for (int i = 0; i <k; i++)
{
for (int j = sum/2; j >=val[i]; j--)//这里不能改为for (int j = val[i]; j <=sum/2; j++)
{
dp[j] = max(dp[j ], dp[j - val[i]] + val[i]);
}
}
cout << "Collection #" << ca++ << ":" << endl;
if (dp[sum / 2] == sum / 2)
cout << "Can be divided." << endl << endl;
else
cout << "Can't be divided." << endl << endl;
}
}
}